# Explain this circuit please?

1. Sep 8, 2016

### cjm181

1. The problem statement, all variables and given/known data

i am working through this question, the final bit i am struggling with, describing what each bit of the circuit does. tying to understand what is going on. Looking at fig 1b

I have identified a bridge rectifier, and the zener diode. I am reading up on these so i am not looking for help with these at the mo! What i cant seem to sus is what the two resistors are doing in the ac supply before the bridge rectifier. And also what is the resistor after the rectifier, the vertical one? The coupling is an optoisolator. Can anyone help me or point me to some further reading please?

2. Relevant equations

3. The attempt at a solution
Are the two resistors to limit the current going into the rectifer?

Thanks

2. Sep 8, 2016

### jaus tail

I don't think the resistors would be to limit the current going into the rectifier. Since the resistors are permanently in the circuits there'd be power loss across them.
Also if you want to reduce current, maybe use one resistor in series why both?

3. Sep 8, 2016

### Numbskull

If you were using resistors to limit the current, you could use two smaller resistors to dissipate the power rather than one big one? Cost / size constraints?

4. Sep 8, 2016

### Staff: Mentor

It could be for current limiting safety. It would appear that the circuit is to be directly interfaced to the mains, and you definitely don't want any direct contact with a mains line to be made directly accessible to external contact as might occur at the coupling interface (say if there's a connector there). An isolation transformer would be preferable, but presumably is omitted to save size, weight, and cost.

5. Sep 8, 2016

### Staff: Mentor

I'd draw in the relevant side of the optocoupler circuit, so it can be considered in association with your 230V components.

6. Sep 8, 2016

### cjm181

its says breifly explain! maybe the resistors are for something to the left not shown?

7. Sep 8, 2016

### Nidum

It could be based on an old style line dropper circuit with bits added on to drive the opto coupler .

The two resistors to the left of the diode bridge drop the voltage across the diode bridge to a level much lower than mains level . The other resistor provides a ballast load to set the current flowing to a level which gives the right voltage drops across the line dropper resistors .

nb: Current taken by opto coupler << current in ballast resistor .

Not a good or an efficient arrangement but very low cost compared with a transformer circuit .

Last edited: Sep 8, 2016
8. Sep 9, 2016

### cjm181

Thanks everyone!

My interpretation of the circuit so far (bear in mind i am a mechanical guy!)

From left to right

The two resistors in the 230V ac supply - to reduce the voltage and current going in to the next component

The diode bridge rectifier - to convert ac to dc

The parallel resistor - to further reduce the current and voltage going to the zener diode

The zener diode - this regulates and stabilises voltage going to the load (in this case the opto isolator). it does this by operating in the breakdown region, where voltage remains relatively constant regardless of applied current (within limits). Also keeps a consistant voltage if the load changes. However not sure how the load would change with an optoisolator?

the parallel diode - not sure about this one. I am going to guess if anything goes wrong with the rectifier, it allows current to flow thru it from bottom to top. but wont allow current to flow from top down, so in that instance all the current / voltage goes to the opto isolator.

Am i miles off?

Thanks
Craig

9. Sep 9, 2016

### Staff: Mentor

Your explanation is okay up until here. The zener here is not used how you may think, and it does not stabilise the voltage across any element (apart from itself!).

I conclude that you haven't followed my earlier advice, then?

10. Sep 9, 2016

### cjm181

ok, so to draw in the optoisolator will be to draw in an LED on the left side of the coupler box, and a receiver on the right side of the coupler box.

can you expand on the comment 'so it can be considered with the 230v components'?

ummm

so my led knowledge is not great, but i know u need to limit the current going thru them, with a resistor!

11. Sep 9, 2016

### Nidum

+1

12. Sep 9, 2016

### Staff: Mentor

Everything between the 230VAC plug and the optoisolator LED/s is there for the purpose of safely providing an appropriate drive current to power those LEDS. You need to consider the whole lot in combination when trying to understand what purpose each element serves, and ask yourself questions about each.

Start by drawing a large sinusoid, superimposing on it the rectified current waveform, then for each element work out what its current waveform will be doing throughout the mains cycle.

13. Sep 9, 2016

### cjm181

ok, not sure then, let me have a think...

the zener, are we calling this a unidirectional zener? based on its symbol.

So, i am calling the top horizontal line positive, and the bottom line negative.

lets say there is a 12v DC supply, and the coupler needs 5v supply, if the zener diode is 7v, wont 7v be lost across the zener as heat, and the coupler will receive 5v across it?

14. Sep 9, 2016

### Staff: Mentor

In this hypothetical circuit you have thrown in for discussion....yes.

15. Sep 9, 2016

### cjm181

ok, back up to post 5

16. Sep 9, 2016

### cjm181

so, diode 1, the zener, allows current and voltage to flow from left to right once the voltage exceeds the knee voltage. an amount of voltage will be lossed as heat. However it does allow large changes in current to pass thru??

so, looking at the led diode in the coupler, and indeed diode 2, what is limiting the current going thru the circuit?

and whats D2 doing? thats going to allow voltage / current to travel upwards, but stop it traveling downwards

17. Sep 9, 2016

### Nidum

Deleted

Last edited: Sep 9, 2016
18. Sep 9, 2016

### Staff: Mentor

Yes. (Though don't overstate "voltage lost as heat" because heat is also a function of current; just need to explain it as a voltage drop across the zener.)

That's a good point---LEDs do need their current limited. Nothing obvious here in your sketch. Perhaps something you have omitted to draw, something between here and the 230V plug?

19. Sep 9, 2016

### cjm181

ok, so to summarise there are 4 things,

1. whats limiting the current going through the circuits for the led and D2
2. Whats the zener doing?
3. what is r1 doing
4. whats d2 doing.

so, answering point 1 - well the bridge cant be limiting current, as that is made of diodes. the only other thing is the two resistors R2 and R3, in the ac line. So R2 and R3 limit the current going thru the circuit!

2, still no closer to this. If the two resistors in the ac side are handling current, then the zener cant be current related. (although i know they let large currents thru. How ever when i type zener diode in to google, alot of the returns are regarding voltage regulation. Further on this, the voltage in the circuit must be controlled, so i am thinking it is still involved in supply to the coupler. i think its working in conjuction with r1.

3?

4. the bridge rectifier is made up of diodes. diodes of this type let voltage / current flow one way, and not the other way (ideally). I know in reality there is a little bit of leakage. I also know that if enough current / voltage is applied in the wrong direction, the diode will start to conduct, and damage it. So if this was to happen, current could flow along the negative line, so what would happen then? Current would travel thru r1, some thru d2, as thats the direction it will work in, but the coupler will become open circuit. if r1 and d2 were not there, then the coupler would be damaged, as the same situation would apply as what happened to the rectifier diodes, the current would become so high the led would start to conduct.

Going the right way, r1 is still in use, the zenner is being used, but d2 (assuming the current / voltage is ok) is not being used.

20. Sep 9, 2016

### Staff: Mentor

That looks okay.

You should be able to determine ballpark values for R2 and R3 by looking at manufacturers' datasheets to find a typical value of current needed to drive an optoisolator LED.

Have you thought about the waveshape being provided to the logic circuits in the PLC? You know what the rectified sinewave looks like.....and the circuit you are examining here manipulates that mains sinewave into becoming what the PLC needs. What waveform would that typically look like, in order to clock the PLC?

Last edited: Sep 9, 2016